Given equation of hyperbola is $9y^2 - 4x^2$ = 36 i.e., $\frac{9y^2}{36} - \frac{4x^2}{36}$ = 1 ⇒ $\frac{y^2}{4} - \frac{x^2}{9}$ = 1 , which is of the form $\frac{y^2}{a^2} - \frac{x^2}{b^2}$ = 1. The foci and vertices of the hyperbola lie on y-axis. Now, $a^2$ = 4 ⇒ a = 2 and $b^2$ = 9 ⇒ b = 3 Also, $c^2$ = $a^2 + b^2$ = 4 + 9 = 13 ⇒ c = $\sqrt{13}$ ∴ Coordinates of foci are (0, ± c) i.e. (0, ± $\sqrt{13}$) ∴ Coordinates of vertices are (0, ±a) i.e. (0, ±2) Eccentricity (e) = $\frac{c}{a}$ = $\frac{\sqrt{13}}{2}$ Length of latus rectum = $\frac{2b^2}{a}$ = $\frac{2 \times 9}{2}$ = 9.